Question

A car moving with a speed of 40 m/s is brought to rest be application of brakes in 400 m.Calculated the acc. and time taken

Answer 1

$\bf\;\underline{Question}$

A car moving with a speed of 40 m/s is brought to rest by application of brakes in 400 m.Calculate the acceleration and time taken

$\bf\;\underline{Explanation}$

• Initial velocity(u)=40 m/s
• Final velocity(v)=0 m/s     (SINCE THE CAR STOPS)
• Distance(S)=400 m
• Acceleration(a)=??
• Time taken(t)=??

FORMULA USED:-

$\longrightarrow\bigstar\bf\;\boxed{\bf\;v^2=u^2+2as}\bigstar$

$\longrightarrow\bigstar\boxed{\bf\;v=u+at}\bigstar$

LET'S BEGIN !

First let's calculate the acceleration using 3rd equation:-

→0²=(40)²+2a(400)

→0=1600+800a

→-1600=800a

→a=1600/800

→a=-2 m/s

So the acceleration is 2 m/s

Now let's calculate the distance travelled

→v=u+at

→0=40+(-2)t

→0=40-2t

→2t=40

→t=20

So we can conclude:-

Acceleration= 2 m\s²

Time taken=20 s

Answer 2

Given

• Initial Velocity (u) = 40 m/s
• Final Velocity (v) = 0 m/s
• Distance covered = 400 m

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To Find

(i) Acceleration (a)

(ii) Time Taken (t)

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Solution

(i) We'll apply the third equation of motion to find the Acceleration of the car.

3rd Equation of Motion → v² - u² = 2as

Substitute the values of varaibles and find the value of 'a'.

⇒ (0)² -  (40)² = 2a(400)

⇒ 0 - 1600 = 800a

⇒ 800a = -1600

⇒ a = -1600 ÷ 800

⇒ a = -2 m/s²

∴ The acceleration of the car is -2 m/s²

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(ii) To find the time taken, we'll apply the 1st Equation of Motion.

1st Equation of Motion → v = u + at

Substitute the values of varaibles and find the value of 't'.

⇒ 0 = 40 + -2(t)

⇒ 40 + -2t = 0

⇒ -2t = -40

⇒ t = -40 ÷ -2

⇒ t = 20 sec

∴ The time taken is 20 seconds.

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